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Twenty-nine percent of americans say they are confident that passenger trips to the moon will occur in their lifetime. You randomly sample 200 americans and ask if they believe passenger trips to the moon will occur in their lifetime. What is the probability that less than 40% of the people sampled will answer yes to the question?

Answer :

lyndalau86

Answer:

The probability is 0.0014

Step-by-step explanation:

When we have two possible outcomes p, q and we repeat the experiment multiple times, we apply the Binomial distribution, in this distribution the probability of getting exactly k successes in n trials is given by

[tex]P(X=k)  = nCk(p)^k(1-p)^{n-k}[/tex]

where p represents the probability of success. nCk refers to the combinations of k elements out of n elements.

In this case, p= .29 (the probability that the person says "yes" being confident that passenger trips to the moon will occur in their lifetime.

q = .71 (The probability that the person says "no")

We sample 200 americans, therefore, we have n = 200 and we need the probability that less than 40% of the people sampled answer "yes".

The 40% of 200 is 80. So, in other words, we need less than 80 people to answer yes to the question.

P(x<80) = P(x=0) + P(x=1) + P(x=2) + .... + P(x=79) = 0.00143466294 (we can use a Binomial Distribution calculator to compute this result)

Therefore, P(x < 80) = 0.0014

Using the normal distribution and the central limit theorem, it is found that there is a 0.9997 = 99.97% probability that less than 40% of the people sampled will answer yes to the question.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
  • By the Central Limit Theorem, for a proportion p in a sample of size n, the sampling distribution of sample proportion is approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex], as long as [tex]np \geq 10[/tex] and [tex]n(1 - p) \geq 10[/tex].

In this problem:

  • Twenty-nine percent of Americans say they are confident that passenger trips to the moon will occur in their lifetime, hence p = 0.29.
  • 200 Americans are sampled, hence n = 200.

The mean and the standard error are given by:

[tex]\mu = p = 0.29[/tex]

[tex]s = \sqrt{\frac{p(1 - p)}{n}} = \sqrt{\frac{0.29(0.71)}{200}} = 0.0321[/tex]

The probability that less than 40% of the people sampled will answer yes to the question is the p-value of Z when X = 0.4, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{0.4 - 0.29}{0.0321}[/tex]

[tex]Z = 3.43[/tex]

[tex]Z = 3.43[/tex] has a p-value of 0.9997.

0.9997 = 99.97% probability that less than 40% of the people sampled will answer yes to the question.

To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213

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